Question

Find the sign of the expression if the terminal point determined by $t$is in the given quadrant.

$\sin t\cos t,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Quadrant\hspace{0.17em}\hspace{0.17em}II}$

Short answer

In quadrant II, $\mathbf{sin}\mathit{t}\mathbf{}\mathbf{cost}\mathbf{<}\mathbf{0}$.

Step-by-step solution

Step 1. State the definition of the terminal point on the unit circle.

Starting at the point $(1,0)$, if $t\ge 0$, then t is the distance along the unit circle in a clockwise direction and if $t<0$, then $\left|t\right|$ is the distance along the unit circle in a clockwise direction. Here, t is a real number and this distance generates a point $P\left(x,y\right)$ on the unit circle. The point $P\left(x,y\right)$ obtained in this way is called the terminal point determined by the real number $t$.

Step 2. Find the sign-in quadrant II. 

Determine the sign of the trigonometric function of $t$by noting the quadrant in which the terminal point lies. The terminal point of $t$ lies in quadrant II and in quadrant II $x$-coordinate is negative and $y$-coordinate is positive.

Step 3. Find the sign of the expression.

Note that for the given point t on the unit circle $\cos t=x$, $\sin t=y$and $\tan t=\frac{y}{x}$where $\left(x,y\right)$is the terminal point to a real number $t$.

Since $x<0$, therefore, $\cos t<0$and because $y>0$, therefore, $\sin t>0$.

$$\begin{gathered} \sin t\cos t=xy \\ <0 \end{gathered}$$

Hence, in quadrant II, $\sin t\cos t<0$.